What is the integral of 6/((x-4)(x+2)) ?

The integral of 6/((x-4)(x+2)) is ln|x-4|-ln|x+2|+C

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integral of 6/((x-4)(x+2))

\int \frac{6}{( x - 4 ) ( x + 2 )} d x

ln ∣ x - 4 ∣ - ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{6}{( x - 4 ) ( x + 2 )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{1}{( x - 4 ) ( x + 2 )} d x

Take the partial fraction of

\frac{1}{( x - 4 ) ( x + 2 )} : \frac{1}{6 ( x - 4 )} - \frac{1}{6 ( x + 2 )}

= 6 \cdot \int \frac{1}{6 ( x - 4 )} - \frac{1}{6 ( x + 2 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 6 (\int \frac{1}{6 ( x - 4 )} d x - \int \frac{1}{6 ( x + 2 )} d x)

\int \frac{1}{6 ( x - 4 )} d x = \frac{1}{6} ln ∣ x - 4 ∣

\int \frac{1}{6 ( x + 2 )} d x = \frac{1}{6} ln ∣ x + 2 ∣

= 6 (\frac{1}{6} ln ∣ x - 4 ∣ - \frac{1}{6} ln ∣ x + 2 ∣)

Simplify

6 (\frac{1}{6} ln ∣ x - 4 ∣ - \frac{1}{6} ln ∣ x + 2 ∣) : ln ∣ x - 4 ∣ - ln ∣ x + 2 ∣

= ln ∣ x - 4 ∣ - ln ∣ x + 2 ∣

Add a constant to the solution

= ln ∣ x - 4 ∣ - ln ∣ x + 2 ∣ + C

y^{^{''}} + 81 y = sec^{2} (9 x)

(\frac{1}{4}, \frac{7}{8}), (4 \frac{1}{4}, 2 \frac{7}{8})

n = 0 \sum \infty \frac{1}{( 2 n + 1 ) ^{6}}

y d x + (1 + x) d y = 0

\frac{\partial}{\partial x} (\frac{x ^{2} + y ^{2} - 7}{x y})

What is the integral of 6/((x-4)(x+2)) ?
The integral of 6/((x-4)(x+2)) is ln|x-4|-ln|x+2|+C